axis, so as to take an infinite number of positions. Hence, we infer that an infinite number of planes may be passed through a given line.

Through three points, not in the same straight line, one plane can be passed, and only one.

Let A, B, and C be the three points: then can one plane be passed through them, and only one. Join two of the points, as A and B, by the line AB. Through AB let a plane be passed, and let this plane be turned around AB until it contains the point C; in this position it will pass through the three points A, B, and C. If now, the plane be turned

B

about AB, in either direction, it will no longer contain the point Chence, one plane can always be passed through three points, and only one; which was to be proved.

Cor. 1. Three points, not in a straight line, determine the position of a plane, because only one plane can be passed through them.

Cor. 2. A straight line and a point without that line, determine the position of a plane, because only one plane can be passed through them.

Cor. 3. Two straight lines which intersect, determine the position of a plane. For, let AB and AC intersect at A then will either line, as AB,

and one point of the

other, as C, determine the position of a plane.

Cor. 4. Two parallel lines determine the position of a

plane. For, let AB and CD be parallel. By definition (B. I., D. 16) two parallel lines always lie in the same plane. But either line, as AB, and any point

of the other, as F, determine the position of a plane hence, two parallels determine the position of a plane.

:

A

-B

C

D

F

PROPOSITION III. THEOREM.

The intersection of two planes is a straight line.

Let AB and CD be two planes: then will their intersection be a straight line.

For, let E and F be any two points common to the planes; draw the straight line EF. This line having two points in the plane AB, will lie wholly in that plane; and having two points in the plane CD,

D

B

E

A

will lie wholly in that plane: hence, every point of EF is common to both planes. Furthermore, the planes can have no common point lying without EF, otherwise there would be two planes passing through a straight line and a point lying without it, which is impossible (P. II., C. 2); hence, the intersection of the two planes is a straight line; which was to be proved.

PROPOSITION IV. THEOREM.

If a straight line is perpendicular to two straight lines at their point of intersection, it is perpendicular to the plane of those lines.

Let MN be the plane of the two lines BB, CC, and let AP be perpendicular to these lines at P: then will

AP be perpendicular to every line of the plane which passes through P, and consequently, to the plane itself.

For, through P, draw in the plane MN, any line PQ; through any point of this line, as Q, draw the line BC, SO that BQ shall be equal to QC (B. IV., Prob. V.); draw AB, AQ, and AC.

A

P

B

M

The base BC, of the triangle BPC, being bisected at Q, we have (B. IV., P. XIV.),

In like manner, we have, from the triangle ABC,

AC2 + AB2 2A Q2 + 2 QC2.

Subtracting the first of these equations from the second, member from member, we have,

2

AC2 - PC2 + AB2 PB2 = 2AQ2 — 2.PQ3.

But, from Proposition XI., C. 1, Book IV., we have,

AP2 =

2AP2 = 2AQ2 — 2PQ2;

A Q2 – P Q2 ; or, AP2 + PQ2 AQ2.

The triangle APQ is, therefore, right-angled at P (B. IV., P. XIII., S.), and consequently, AP is perpendicular to PQ hence, AP is perpendicular to every line of the plane MN passing through P, and consequently, to the plane itself; which was to be proved.

A

N

Cor. 1. Only one perpendicular can be drawn to a plane from a point without the plane. For, suppose two perpendiculars, as AP and AQ, could be drawn from the point A to the plane MN. Draw PQ; then the triangle APQ would have two right angles, APQ and AQP; which is impossible (B. I., P. XXV., C. 3).

P

M

Cor. 2. Only one perpendicular can be drawn to a plane from a point of that plane. For, suppose that two perpendiculars could be drawn to the plane MN, from the point P. Pass a plane through the perpendiculars, and let PQ be its intersection with MN; then we should have two perpendiculars drawn to the same straight line from a point of that line; which is impossible (B. I., P. XIV., C.).

PROPOSITION V. THEOREM.

If from a point without a plane, a perpendicular be drawn to the plane, and oblique lines be drawn to different points of the plane:

1o. The perpendicular will be shorter than any oblique line:

2°. Oblique lines which meet the plane at equal distances from the foot of the perpendicular, will be equal:

3.° Of two oblique lines which meet the plane at unequal distances from the foot of the perpendicular, the one which meets it at the greater distance will be the longer.

Let A be a point without the plane MN; let AP be perpendicular to the plane; let AC, AD, be any two oblique lines meeting the plane at equal distances from the foot of the perpendicular; and let AC and AE be any

two oblique lines meeting the plane at unequal distances from the foot of the perpendicular:

2o. AC and AD will be equal.

For, draw PD; then the right-angled triangles APC, APD, will have the side AP common, and the sides PC, PD, equal: hence, the triangles are equal in all their parts, and consequently, AC and AD will be equal; which was to be proved.

3o. AE will be greater than AC.

For, draw PE, and take PB equal to PC; draw AB: then will AE be greater than AB (B. I., P. XV.); but AB and AC are equal: hence, AE is greater than AC; which was to be proved.

Cor. The equal oblique lines AB, AC, AD, meet the plane MN in the circumference of a circle, whose centre is P, and whose radius is PB: hence, to draw a perpendicular to a given plane MN, from a point A, without that plane, find three points B, C, D, of the plane equally distant from A, and then find the centre P, of the circle whose circumference passes through these points: then will AP be the perpendicular required.

Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN. The equal oblique lines AB, AC, AD, are all equally inclined to the plane MN. The inclination of AE is less than the inclination of any shorter line AB.